Work, Energy and Power

Class 10 Physics • Chapter 02 (Deep Detail)

1. Work

Work is said to be done only when a force applied on a body makes the body move (i.e., there is displacement).

FORMULA

$$ W = F \times S $$

If force acts at an angle $\theta$ to the direction of displacement:

$$ W = F S \cos \theta $$

SI Unit: Joule (J) | CGS Unit: Erg

Relation: $1 \text{ joule} = 10^7 \text{ erg}$

Special Cases:

$\theta = 0^\circ$ (Max Work): Force and displacement in same direction. $W = F \times S$. (e.g., Free fall).
$\theta = 90^\circ$ (Zero Work): Force is perpendicular to displacement. $W = 0$. (e.g., Coolie carrying load on horizontal ground, Circular motion).
$\theta = 180^\circ$ (Negative Work): Force opposes displacement. $W = - F \times S$. (e.g., Friction).

Practice Q1: Work with Angle

[IMAGE PLACEHOLDER: FORCE AT ANGLE]

Block on horizontal surface. Force F=10N acts at angle 60 degrees to horizontal. Displacement S=2m. Show components. (Invert in dark mode).

NUMERICAL A boy pulls a toy car with a force of 10 N at an angle of $60^\circ$ to the horizontal. Calculate work done in moving it by 2 m.

Solution: $W = F S \cos \theta = 10 \times 2 \times \cos 60^\circ = 20 \times 0.5 = 10 \text{ J}$.

Practice Q2: Work Against Gravity

NUMERICAL A coolie lifts a load of 20 kg from the ground to a height of 1.5 m. Calculate the work done by him on the load. (Take $g = 10 \text{ m/s}^2$)

Solution: $W = mgh = 20 \times 10 \times 1.5 = 300 \text{ J}$.

ZERO WORK ALERT A satellite revolving around Earth does ZERO WORK because the centripetal force is always perpendicular ($90^\circ$) to the instantaneous direction of motion (tangent).

Topic Check: Zero Work

Q: A coolie holding a heavy suitcase stands still for 5 minutes. Work done?

Ans: Zero. (No displacement, $S=0$).

2. Power

The rate of doing work is called Power.

$$ P = \frac{\text{Work Done (W)}}{\text{Time Taken (t)}} $$

Since $W = F \times S$, we can also write:

$$ P = F \times v \quad (\text{Since } v = S/t) $$

SI Unit: Watt (W) or J/s

Horsepower (hp): $1 \text{ hp} = 746 \text{ W}$

Commercial Unit of Energy (Electricity)

Kilowatt-hour (kWh): The energy consumed by an appliance of power 1 kW in 1 hour.

Conversion: $1 \text{ kWh} = 1 \text{ kW} \times 1 \text{ h}$

$1 \text{ kWh} = 1000 \text{ W} \times 3600 \text{ s} = 3.6 \times 10^6 \text{ J}$

Topic Check: Power

Q: If it takes 1 minute to lift a 100 N weight to 5 m height, calculate power.

Ans: $P = W/t = (100 \times 5) / 60 = 500/60 = 8.33 \text{ W}$.

Practice Q4: Power Comparison

CONCEPTUAL Two students A and B of equal mass climb up a staircase. A takes 20 s and B takes 25 s. Ratio of work done and power developed?

Solution:

1. Work Done: Both have same mass and climb same height. So, $W_A = W_B = mgh \implies \text{Ratio } 1:1$.

2. Power: $P \propto 1/t$. Ratio $P_A : P_B = t_B : t_A = 25 : 20 = 5 : 4$.

Practice Q5: Horsepower Logic

NUMERICAL An electric motor of power 2 HP operates for 5 hours daily. Calculate the energy consumed in kWh in 10 days.

Solution:

1. Power: $2 \text{ HP} = 1.492 \text{ kW}$.

2. Time: $5 \times 10 = 50 \text{ h}$.

3. Energy: $E = P \times t = 1.492 \times 50 = 74.6 \text{ kWh}$.

3. Energy

The capacity to do work is called Energy. It is a scalar quantity. SI unit is Joule.

A. Potential Energy (U)

Energy possessed by a body due to its position or configuration (shape).

Gravitational PE: Due to height.
Elastic PE: Due to change in shape (stretched bow, compressed spring).

GRAVITATIONAL PE

$$ U = mgh $$

Where $m$ = mass, $g$ = acceleration due to gravity, $h$ = height above ground.

Work done against gravity stored as potential energy.

B. Kinetic Energy (K)

Energy possessed by a body due to its motion.

KINETIC ENERGY

$$ K = \frac{1}{2} mv^2 $$

Forms of KE: Translational (straight line), Rotational (spinning), Vibrational.

Relationship between KE and Momentum (p)

Since $p = mv$,

$$ K = \frac{p^2}{2m} \quad \text{or} \quad p = \sqrt{2mK} $$

(Very important for numericals comparing two bodies)

Practice Q6: Light vs Heavy Body

REASONING A light body and a heavy body have the same momentum. Which one has greater kinetic energy?

Ans: Light body.

Reason: $K = p^2 / 2m$. Since $p$ is constant, $K \propto 1/m$. The body with smaller mass ($m$) will have larger KE.

Practice Q7: Kinetic Calculation

NUMERICAL If the kinetic energy of a body is 2500 J and its momentum is 500 kg m/s, find its mass.

Solution: $m = \frac{p^2}{2K} = \frac{500^2}{2 \times 2500} = \frac{250000}{5000} = 50 \text{ kg}$.

Work-Energy Theorem

The work done by a force on a moving body is equal to the increase in its kinetic energy.

$$ W = K_f - K_i $$ $$ = \frac{1}{2}mv^2 - \frac{1}{2}mu^2 $$

4. Conservation of Mechanical Energy

[IMAGE PLACEHOLDER: ROLLER COASTER ENERGY]

Diagram of a simple pendulum or roller coaster track. Label 'Max PE' at the highest points and 'Max KE' at the lowest points. Show energy conversion flow.

The Principle of Conservation of Energy states that energy can neither be created nor destroyed, only changed from one form to another.

Mechanical Energy (E) = Potential Energy (U) + Kinetic Energy (K)

For a freely falling body (ignoring air resistance), the total mechanical energy remains constant at all points.

Verification (Freely Falling Body)

Consider a body of mass $m$ at height $H$.

At Top (Position A): $v=0$.
$K=0$. $U=mgH$.
Total = $mgH$.
At Middle (Position B - fall distance $x$):
$U = mg(H-x)$.

Using $v^2 = u^2 + 2gx \implies v^2 = 2gx$,

so $K = \frac{1}{2}m(2gx) = mgx$.
Total = $mg(H-x) + mgx = mgH$.
At Bottom (Position C): $h=0$.
$U = 0$.
Using $v^2 = 2gH$, so $K = \frac{1}{2}m(2gH) = mgH$.
Total = $mgH$.

Conclusion: Total mechanical energy is conserved throughout the fall.

Practice Q8: Conservation Pendulum

[IMAGE PLACEHOLDER: PENDULUM POSITIONS A B C]

Simple pendulum. A (mean), B (extreme). Vertical height difference 'h'. Mass m. (Invert in dark mode).

NUMERICAL A simple pendulum of mass 200g swings from A to B. If the vertical height of B above A is 10cm, find its speed at A. ($g=10$).

Solution: Gain in KE at A = Loss in PE at B.

$\frac{1}{2} mv^2 = mgh \implies v = \sqrt{2gh}$.

$v = \sqrt{2 \times 10 \times 0.1} = \sqrt{2} = 1.414 \text{ m/s}$.

Practice Q9: Energy Conversion

THINKING A ball is thrown vertically upwards. Discuss the change in KE and PE.

Ans: As it rises, $v$ decreases ($\downarrow$ KE) and $h$ increases ($\uparrow$ PE). At max height, KE=0, PE is Max. Total Energy is constant.

Energy Transformations

Electric Motor: Electric $\to$ Mechanical.
Dynamo/Generator: Mechanical $\to$ Electric.
Loudspeaker: Electric $\to$ Sound.
Microphone: Sound $\to$ Electric.
Battery (Cell): Chemical $\to$ Electric.
Photosynthesis: Light $\to$ Chemical.

BOARD TIPS

Remember: Power is scalar, but depends on time. Work does not depend on time. Two boys doing same work in different time have different power.
If speed is doubled, Kinetic Energy becomes 4 times ($v^2$ dependency).
Always check units! Convert mass to kg, distance to metres, time to seconds before calculating Joules/Watts.
Work by Gravity: When body falls, force and displacement are in same direction ($W = +mgh$). When thrown up, they are opposite ($W = -mgh$).